hdu4282 A very hard mathematic problem

A very hard mathematic problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3017 Accepted Submission(s): 875

Problem Description

　　Haoren is very good at solving mathematic problems. Today he is working a problem like this:
　　Find three positive integers X, Y and Z (X < Y, Z > 1) that holds
　　 X^Z + Y^Z + XYZ = K
　　where K is another given integer.
　　Here the operator “^” means power, e.g., 2^3 = 2 * 2 * 2.
　　Finding a solution is quite easy to Haoren. Now he wants to challenge more: What’s the total number of different solutions?
　　Surprisingly, he is unable to solve this one. It seems that it’s really a very hard mathematic problem.
　　Now, it’s your turn.

Input

　　There are multiple test cases.
　　For each case, there is only one integer K (0 < K < 2^31) in a line.
　　K = 0 implies the end of input.
　　

Output

　　Output the total number of solutions in a line for each test case.

Sample Input

   
   
   
   

    
    
    
    9
53
6
0
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    1
1
0
　　

    
    
    
    
 
     
 
      Hint 
     
 9 = 1^2 + 2^2 + 1 * 2 * 2 53 = 2^3 + 3^3 + 2 * 3 * 3 
    
    
    
    
     
   
   
   
   

Source

2012 ACM/ICPC Asia Regional Tianjin Online

Recommend

liuyiding

暴力就可以过！

#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
__int64 pow(__int64 a,__int64 b)
{
    __int64 i,ans=1;
    for(i=0;i<b;i++)
    {
        ans*=a;
    }
    return ans;
}
int main()
{
   __int64 k,ans,a1,a2,sum;
   __int64 x,y,z;
   while(scanf("%I64d",&k)!=EOF&&k)
   {
       ans=0;
        __int64 temp =(__int64) sqrt(k);
       if(temp*temp==k)//加这里，就是250ms,不加，就超时了！
            ans += (temp-1)>>1;
       for(z=3;z<31;z++)
       {
           for(x=1;;x++)
           {
               a1=pow(x,z);
               if(a1>k/2)
               break;
               for(y=x+1;;y++)
               {
                   a2=pow(y,z);
                   if(a2>k)
                   break;
                   sum=0;
                   sum=a1+a2+x*y*z;
                   if(sum>k)
                   break;
                   if(sum==k)
                   {
                       ans++;
                       break;
                   }
               }
           }
       }
       printf("%I64d\n",ans);
   }
    return 0;
}